CBSE Sample Paper 2025-26 · Section E · 4 marks

Q37: Case study: profit from energy-efficient LED bulbs

Sample 2025-26Ch 6 · Application of Derivatives

The question

Case Study -2

LED bulbs are energy-efficient because they use significantly less electricity than traditional bulbs while producing the same amount of light. They convert more energy into light rather than heat, reducing waste. Additionally, their long lifespan means fewer replacements, saving resources and money over time.

A company manufactures a new type of energy-efficient LED bulb. The cost of production and the revenue generated by selling $x$ bulbs (in an hour) are modelled as

$C(x) = 0.5x^2 - 10x + 150$ and $R(x) = -0.3x^2 + 20x$ respectively, where $C(x)$ and $R(x)$ are both in ₹.

To maximize the profit, the company needs to analyze these functions using calculus. Use the given models to answer the following questions:

I. Derive the profit function $P(x)$ (1 mark)

II. Find the critical points of $P(x)$ ⁠. (1 mark)

III A. Determine whether the critical points correspond to a maximum or a minimum profit by using the second derivative test. (2 marks)

III B. Identify the possible practical value of $x$ (i.e., the number of bulbs that can realistically be produced and sold) that can maximize the profit, if the resources available and the expenditure on machines allows to produce minimum 10 but not more than 18 bulbs per hour. Also calculate the maximum profit. (2 marks)

Exam answer

\begin{aligned} P(x) &= R(x) - C(x) \\ &= -0.3x^2 + 20x \\ &\quad - (0.5x^2 - 10x + 150) \\ &= -0.3x^2 - 0.5x^2 \\ &\quad + 20x + 10x - 150 \end{aligned}

Why?

Profit is what is left of the revenue after paying the cost: $P = R - C$ ⁠.

Watch out: the minus sign changes the sign of every term of $C(x)$ ⁠:

\begin{aligned} &-(0.5x^2 - 10x + 150) \\ &= -0.5x^2 + 10x - 150 \end{aligned}

Then combine like terms: $-0.3 - 0.5 = -0.8$ and $20 + 10 = 30$ ⁠.

$P(x) = -0.8x^2 + 30x - 150$

II. $P'(x) = -1.6x + 30$ ⁠. For critical points, $P'(x) = 0$ ⁠:

Why?

Differentiate term by term: $\frac{d}{dx}(x^2) = 2x$ ⁠, so $-0.8x^2$ gives $-1.6x$ ⁠; $30x$ gives 30; the constant gives 0.

\begin{aligned} P'(x) &= -0.8(2x) + 30 - 0 \\ &= -1.6x + 30 \end{aligned}

A critical point is where the derivative is zero: the graph is momentarily flat there.

\begin{aligned} -1.6x + 30 &= 0 \\ \Rightarrow x &= \frac{30}{1.6} = \frac{300}{16} \end{aligned}

Why?

Multiply the top and the bottom by 10 to remove the decimal. Then $\frac{300}{16} = \frac{75}{4} = 18.75$ ⁠.

\begin{aligned} -1.6x + 30 &= 0 \\ 30 &= 1.6x \\ x &= \frac{30}{1.6} = \frac{30 \times 10}{1.6 \times 10} = \frac{300}{16} \\ &= \frac{300 \div 4}{16 \div 4} = \frac{75}{4} \\ &= 18.75 \end{aligned}

The critical point is $x = 18.75$ ⁠.

III A. $P''(x) = -1.6$ ⁠.

Why?

Differentiate $P'(x) = -1.6x + 30$ once more. The derivative of $-1.6x$ is $-1.6$ ⁠, and of 30 is 0.

In particular, $P''(18.75) = -1.6 < 0$ ⁠.

Why?

Second derivative test: at a critical point $c$ (where $P'(c) = 0$ ⁠),

\begin{aligned} P''(c) < 0 &\Rightarrow \text{maximum} \\ P''(c) > 0 &\Rightarrow \text{minimum} \end{aligned}

Here $P''$ is the constant $-1.6$ ⁠, negative for every $x$ ⁠, so its value at 18.75 is negative.

So, the critical value $x = 18.75$ corresponds to a maximum profit.

III B. $x$ is a number of bulbs with $10 \le x \le 18$ ⁠. For these $x$ ⁠,

\begin{aligned} P'(x) &= 30 - 1.6x \\ &\ge 30 - 1.6 \times 18 \\ &= 1.2 > 0 \end{aligned}

Why?

The best point 18.75 is not allowed: at most 18 bulbs can be made, and the number of bulbs is a whole number.

$30 - 1.6x$ gets smaller as $x$ grows, so on $[10, 18]$ it is smallest at $x = 18$ ⁠: $1.6 \times 18 = 28.8$ and $30 - 28.8 = 1.2$ ⁠.

\begin{aligned} 1.6 \times 18 &= \frac{16 \times 18}{10} = \frac{288}{10} \\ &= 28.8 \\ 30 - 28.8 &= 1.2 \end{aligned}

$P' > 0$ means $P$ is increasing, so the largest profit in the allowed range is at the right end, $x = 18$ ⁠.

So $P(x)$ is increasing on $[10, 18]$ ⁠, and practically 18 bulbs per hour give the maximum profit.

\begin{aligned} P(18) &= -0.8 \times 18^2 \\ &\quad + 30 \times 18 - 150 \\ &= -259.2 + 540 - 150 \end{aligned}

Why?

$18^2 = 324$ and $0.8 \times 324 = 259.2$ ⁠. Also $30 \times 18 = 540$ ⁠.

\begin{aligned} 18^2 &= 18 \times 18 = 324 \\ 0.8 \times 324 &= \frac{8 \times 324}{10} = \frac{2592}{10} \\ &= 259.2 \\ 30 \times 18 &= 540 \end{aligned}

Then $540 - 150 = 390$ and $390 - 259.2 = 130.8$ ⁠.

$\therefore$ Maximum profit $= P(18) =$ ₹ 130.80, at $x = 18$ bulbs per hour.

Where this comes from

Chapter 6, Application of Derivatives: Maximum profit: second derivative test.

Question type: III A Local maxima and minima; III B Absolute maximum and minimum on a closed interval

How to spot it: click to see

III A Type 7. Local maxima and minima: the second derivative test

Spot it: "Find the local maxima and local minima" or "the local maximum value".

Here: $P(x) = -0.8x^2 + 30x - 150$ has its critical point at $x = 18.75$ ⁠, and $P''(x) = -1.6$ is negative there, so it gives a maximum profit.

III B Type 9. Absolute maximum and minimum on a closed interval

Spot it: an interval with square brackets, like $[0, 3]$ ⁠, and "maximum value" or "absolute maximum".

Here: Now $x$ is held to $[10, 18]$ ⁠, and 18.75 lies outside it. $P'(x) = 30 - 1.6x$ stays positive on the interval, so the largest profit is at the end point $x = 18$ ⁠.

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